tanα=1,tanβ=4,求tan(2α-β)=?
∵tanα=1,∴α=π/4+kπ,2α=π/2+2kπ.
∴tan(2α-β)=tan(π/2+2kπ-β)=tan(π/2-β)=cotβ=1/tanβ=1/4